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M31-22

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New post 14 Jun 2015, 13:15
3
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A
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C
D
E

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Question Stats:

38% (01:34) correct 63% (01:53) wrong based on 56 sessions

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New post 14 Jun 2015, 13:15
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Official Solution:

For prime numbers \(x\) and \(y\), \(x^3*y^5=z^4\). How many positive factors does positive integer \(z\) have?

A. One
B. Two
C. Three
D. Four
E. Five


Notice that \(z^4\) is a perfect square, thus it must have odd number of factors. Since \(z^4\) equals to \(x^3*y^5\), then \(x^3*y^5\) must also have odd number of factors.

Now, if \(x\) and \(y\) are different primes, then the number of factors of \(x^3*y^5\) will be \((3 + 1)(5 + 1) = 24\), which is not odd. Therefore \(x\) and \(y\) MUST be the same prime.

In this case \(x^3*y^5 = x^3*x^5 = x^8 = z^4\).

\(x^8 = z^4\);

\(x^2 = z\).

Since \(x\) is prime, then the number of factors of \(z\) is \((2 + 1) = 3\).


Answer: C
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New post 31 Oct 2015, 18:08
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."



Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?
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New post 01 Nov 2015, 02:24
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HunterJ wrote:
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."



Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?


1. Any integer in even power is a perfect square.
2. \(z^4\) is a perfect square. A perfect square has odd number of factors. Since \(z^4\) equals to \(x^3*y^5\), then \(x^3*y^5\) (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

BACK TO YOUR QUESTION:
In case of a perfect square, say for \(6^2 = 2^2*3^2\), the powers of primes will be always even, so when counting the number of factors we'd have \((even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd\).

Tips about the perfect square:
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

Hope it helps.
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Re: M31-22  [#permalink]

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New post 06 Nov 2016, 19:46
z cannot be a prefect square unless x=y. x and y are prime numbers, we cant get a prefect square with unequal powers of prime numbers. hence, x=y.
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Re M31-22  [#permalink]

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New post 21 May 2017, 23:41
I think this is a high-quality question and I agree with explanation.
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Re: M31-22  [#permalink]

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New post 11 Jun 2017, 12:06
Bunuel wrote:
Official Solution:

For prime numbers \(x\) and \(y\), \(x^3*y^5=z^4\). How many positive factors does positive integer \(z\) have?

A. One
B. Two
C. Three
D. Four
E. Five


Notice that \(z^4\) is a perfect square, thus it must have odd number of factors. Since \(z^4\) equals to \(x^3*y^5\), then \(x^3*y^5\) must also have odd number of factors.

Now, if \(x\) and \(y\) are different primes, then the number of factors of \(x^3*y^5\) will be \((3 + 1)(5 + 1) = 24\), which is not odd. Therefore \(x\) and \(y\) MUST be the same prime.

In this case \(x^3*y^5 = x^3*x^5 = x^8 = z^4\).

\(x^8 = z^4\);

\(x^2 = z\).

Since \(x\) is prime, then the number of factors of \(z\) is \((2 + 1) = 3\).


Answer: C



SO Bunuel just to reiterate last line. Does that mean the factors are 1,x,z ?
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New post 12 Jun 2017, 04:25
mbsingh wrote:
Bunuel wrote:
Official Solution:

For prime numbers \(x\) and \(y\), \(x^3*y^5=z^4\). How many positive factors does positive integer \(z\) have?

A. One
B. Two
C. Three
D. Four
E. Five


Notice that \(z^4\) is a perfect square, thus it must have odd number of factors. Since \(z^4\) equals to \(x^3*y^5\), then \(x^3*y^5\) must also have odd number of factors.

Now, if \(x\) and \(y\) are different primes, then the number of factors of \(x^3*y^5\) will be \((3 + 1)(5 + 1) = 24\), which is not odd. Therefore \(x\) and \(y\) MUST be the same prime.

In this case \(x^3*y^5 = x^3*x^5 = x^8 = z^4\).

\(x^8 = z^4\);

\(x^2 = z\).

Since \(x\) is prime, then the number of factors of \(z\) is \((2 + 1) = 3\).


Answer: C



SO Bunuel just to reiterate last line. Does that mean the factors are 1,x,z ?


Absolutely. Say if x = 2, then y = 4 and the factors of 4 are 1, 2, and 4.
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Re: M31-22  [#permalink]

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New post 21 Jul 2017, 03:33
Bunuel wrote:
HunterJ wrote:
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?


1. Any integer in even power is a perfect square.
2. \(z^4\) is a perfect square. A perfect square has odd number of factors. Since \(z^4\) equals to \(x^3*y^5\), then \(x^3*y^5\) (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

BACK TO YOUR QUESTION:
In case of a perfect square, say for \(6^2 = 2^2*3^2\), the powers of primes will be always even, so when counting the number of factors we'd have \((even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd\).

Tips about the perfect square:
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

Hope it helps.



Hi Bunuel,

Considering the 4th point in the tips for perfect square (great tips btw), we can directly assume in the question that x and y are the same prime numbers since their powers in the question are odd and z is a perfect square? Thanks!
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New post 21 Jul 2017, 03:43
ashikaverma13 wrote:
Bunuel wrote:
HunterJ wrote:
"Notice that z4 is a perfect square, thus it must have odd number of factors. Since z4 equals to x3∗y5, then x3∗y5 must also have odd number of factors."

Question about perfect squares. Would z^6 also be a perfect square?

Also, why do x^3 and y^5 need odd factors, given z^4?


1. Any integer in even power is a perfect square.
2. \(z^4\) is a perfect square. A perfect square has odd number of factors. Since \(z^4\) equals to \(x^3*y^5\), then \(x^3*y^5\) (the product of x^3 and y^5, not the multiples separately) must also have odd number of factors.

Finding the Number of Factors of an Integer:

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

BACK TO YOUR QUESTION:
In case of a perfect square, say for \(6^2 = 2^2*3^2\), the powers of primes will be always even, so when counting the number of factors we'd have \((even + 1)(even + 1) = (2 + 1)(2 + 1) = 3*3 = odd*odd = odd\).

Tips about the perfect square:
1. The number of distinct factors of a perfect square is ALWAYS ODD. The reverse is also true: if a number has the odd number of distinct factors then it's a perfect square;

2. The sum of distinct factors of a perfect square is ALWAYS ODD. The reverse is NOT always true: a number may have the odd sum of its distinct factors and not be a perfect square. For example: 2, 8, 18 or 50;

3. A perfect square ALWAYS has an ODD number of Odd-factors, and EVEN number of Even-factors. The reverse is also true: if a number has an ODD number of Odd-factors, and EVEN number of Even-factors then it's a perfect square. For example: odd factors of 36 are 1, 3 and 9 (3 odd factor) and even factors are 2, 4, 6, 12, 18 and 36 (6 even factors);

4. Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.

Hope it helps.



Hi Bunuel,

Considering the 4th point in the tips for perfect square (great tips btw), we can directly assume in the question that x and y are the same prime numbers since their powers in the question are odd and z is a perfect square? Thanks!

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Yes, that's correct.
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New post 26 Jul 2017, 10:20
Just curious multiplication of 2 fractions can be an integer = 2/3 x 3/2 is equal to 1. Is it not possible that x is not equal to y? we find two prime numbers such that forth root of X^3 and Y^5 yield fractions that are reciprocal to each other. Not possible???



P.S. I solved it by thinking that since the question is directly asking how many Prime factor , and because there can be only one correct ans take both X and Y same ( Try to solve with special unique cases when you can) :-D
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New post 24 Aug 2017, 16:19
I think this is a high-quality question and I agree with explanation.
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New post 03 Oct 2017, 04:21
I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I can't think of any positive integer z which satisfies the above equation. the only way according me which satisfies the equation is when x and y are same.
can explain me with a help of an example?
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New post 03 Oct 2017, 04:55
Amanrohra wrote:
For prime numbers \(x\) and \(y\), \(x^3*y^5=z^4\). How many positive factors does positive integer \(z\) have?

A. One
B. Two
C. Three
D. Four
E. Five


Notice that \(z^4\) is a perfect square, thus it must have odd number of factors. Since \(z^4\) equals to \(x^3*y^5\), then \(x^3*y^5\) must also have odd number of factors.

Now, if \(x\) and \(y\) are different primes, then the number of factors of \(x^3*y^5\) will be \((3 + 1)(5 + 1) = 24\), which is not odd. Therefore \(x\) and \(y\) MUST be the same prime.

In this case \(x^3*y^5 = x^3*x^5 = x^8 = z^4\).

\(x^8 = z^4\);

\(x^2 = z\).

Since \(x\) is prime, then the number of factors of \(z\) is \((2 + 1) = 3\).


Answer: C

I think this is a poor-quality question and the explanation isn't clear enough, please elaborate. I can't think of any positive integer z which satisfies the above equation. the only way according me which satisfies the equation is when x and y are same.
can explain me with a help of an example?


It seems that you did not read the solution carefully.

YES, as it's clearly stated in the solution, x and y MUST be the same prime. For example consider x = y = 2. Note here that unless it is explicitly stated otherwise, different variables CAN represent the same number.
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New post 04 Oct 2017, 01:41
This is a high quality question.
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New post 26 Nov 2018, 03:22
Another approach would be to pick numbers, which worked for me...

\(x=3\) and \(y=3\), the question does not state that the prime number have to be distinct... this makes it easier to work with the same primes.

\(3^3*3^5=3^8\) so we know that \(3^8=z^4\), therefore \(9^4=z^4\) --> \(z=9\)

Factors of \(z=9\) --> 1,3,9
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New post 26 Feb 2019, 10:47
I solved this Q like this:

\(x^3 * y^5 = (x*y)^3*y^2\)

Now y^2 has to be equal to x*y to make it a power of 4.

\(y^2= x*y\) leads to x=y
Therefore same prime numbers.

So \(z= (prime^2)^2\)

Therefore number of factors = 2+1.
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Re: M31-22   [#permalink] 26 Feb 2019, 10:47
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